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Question: If a, b, c, d are non zero real numbers such that \[\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+...

If a, b, c, d are non zero real numbers such that
(a2+b2+c2)(b2+c2+d2)≤(ab+bc+cd)2\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+c^{2}+d^{2}\right) \leq \left( ab+bc+cd\right)^{2} , then a, b, c, d are in
A) AP
B) GP
C) HP
D) none of these

Explanation

Solution

Hint: In this question it is given that if a, b, c, d are non zero real numbers such that
(a2+b2+c2)(b2+c2+d2)≤(ab+bc+cd)2\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+c^{2}+d^{2}\right) \leq \left( ab+bc+cd\right)^{2} , then we have to find the relation of a, b, c and d. So to find the solution we have to expand and rearrange the given inequation by using proper identity.
So the identity that we will be using are,
(x+y+z)2=x2+y2+z2+2xy+2yz+2zx\left( x+y+z\right)^{2} =x^{2}+y^{2}+z^{2}+2xy+2yz+2zx.......(1)
(x−y)2=x2−2xy+y2\left( x-y\right)^{2} =x^{2}-2xy+y^{2}..........(2)
Complete step-by-step solution:
Given,
(a2+b2+c2)(b2+c2+d2)≤(ab+bc+cd)2\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+c^{2}+d^{2}\right) \leq \left( ab+bc+cd\right)^{2}
⇒a2b2+a2c2+a2d2+b4+b2c2+b2d2+c2b2+c4+c2d2≤(ab)2+(bc)2+(cd)2+2(ab)(bc)+2(bc)(cd)+2(cd)(ab)\Rightarrow a^{2}b^{2}+a^{2}c^{2}+a^{2}d^{2}+b^{4}+b^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+c^{4}+c^{2}d^{2}\leq \left( ab\right)^{2} +\left( bc\right)^{2} +\left( cd\right)^{2} +2\left( ab\right) \left( bc\right) +2\left( bc\right) \left( cd\right) +2\left( cd\right) \left( ab\right) [ using formula (1) in RHS]
⇒a2b2+a2c2+a2d2+b4+b2c2+b2d2+c2b2+c4+c2d2≤a2b2+b2c2+c2d2+2acb2+2bdc2+2adbc\Rightarrow a^{2}b^{2}+a^{2}c^{2}+a^{2}d^{2}+b^{4}+b^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+c^{4}+c^{2}d^{2}\leq a^{2}b^{2}+b^{2}c^{2}+c^{2}d^{2}+2acb^{2}+2bdc^{2}+2adbc
Now, cancelling a2b2, b2c2, c2d2a^{2}b^{2},\ b^{2}c^{2},\ c^{2}d^{2} in the both side of the above equation, we get
a2c2+a2d2+b4+b2d2+c2b2+c4≤2acb2+2bdc2+2adbca^{2}c^{2}+a^{2}d^{2}+b^{4}+b^{2}d^{2}+c^{2}b^{2}+c^{4}\leq 2acb^{2}+2bdc^{2}+2adbc
⇒a2c2−2acb2+b4+b2d2−2bdc2+c4+a2d2−2adbc+b2c2≤0\Rightarrow a^{2}c^{2}-2acb^{2}+b^{4}+b^{2}d^{2}-2bdc^{2}+c^{4}+a^{2}d^{2}-2adbc+b^{2}c^{2}\leq 0 [ by changing the side]
⇒(ac)2−2(ac)b2+(b2)2+(bd)2−2(bd)c2+(c2)2+(ad)2−2(ad)(bc)+(bc)2≤0\Rightarrow \\{ \left( ac\right)^{2} -2\left( ac\right) b^{2}+\left( b^{2}\right)^{2} \\} +\\{ \left( bd\right)^{2} -2\left( bd\right) c^{2}+\left( c^{2}\right)^{2} \\} +\\{ \left( ad\right)^{2} -2\left( ad\right) \left( bc\right) +\left( bc\right)^{2} \\} \leq 0
⇒(ac−b2)2+(bd−c2)2+(ad−bc)2≤0\Rightarrow \left( ac-b^{2}\right)^{2} +\left( bd-c^{2}\right)^{2} +\left( ad-bc\right)^{2} \leq 0 [ using formula (2)]
If the summation of the squires of more than 1 term is zero then it implies each term is equal to zero.
i.e, (ac−b2)=0, (bd−c2)=0 and (ad−bc)=0\left( ac-b^{2}\right) =0,\ \left( bd-c^{2}\right) =0\text{ and } \left( ad-bc\right) =0
Therefore, b2=acb^{2}=ac; c2=bdc^{2}=bd; ad=bcad=bc
Hence, a, b, c, d are in G.P.
Therefore, the correct option is option B.
Note: So you might be thinking how we concluded a, b, c, d are in GP just by these obtained expressions b2=acb^{2}=ac; c2=bdc^{2}=bd; ad=bcad=bc, because as we know that if a, b, c, d are in Geometric Progression(GP) then ‘b’ and ‘c’ will be the geometric mean of a, c and b, d respectively, i.e, b2=acb^{2}=ac; c2=bdc^{2}=bd, and they also have common ratio, i.e, ba=cb=dc\dfrac{b}{a} =\dfrac{c}{b} =\dfrac{d}{c}, so from here we can write ad=bcad=bc.